Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Answer:
c is the correct option
Step-by-step explanation:
from,
f'(x) = h >0 f(x + h) - f(x)
h
f(x) = - √2x
f(x + h) = - √(2x + h)
f'(x) = h>0 -√(2x + h) - √2x
h
rationalize the denominator
= h>0 -√(2x + h) + √2x (-√(2x + h) - √2x)
h (-√(2x + h) - √2x)
= h>0 4x + 2h - 4x
h(-√(2x + h) -√2x)
= h>0 2h
h(-√(2x+h) - √2x)
= h>0 2
-√(2x+h) - √2x
[tex]\\ \sf\longmapsto f(x)=\sqrt{2x}[/tex]
Now
[tex]\\ \sf\longmapsto \dfrac{1}{-(\sqrt{2x+2h}-\sqrt{2x})}[/tex]
There we plot 2h because if we break root over then it becomes √2h which satisfies f(x)
[tex]\\ \sf\longmapsto \dfrac{1}{-\sqrt{2x+2h}+\sqrt{2x}}[/tex]
Option D
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.